12+1−3= - Infinity Learn

A
$\frac{(1 + \sqrt{3} - \sqrt{2}) (2 + \sqrt{6})}{4}$
B
$\frac{1 + \sqrt{3} - \sqrt{2}}{- 8}$
C
$\frac{2 + \sqrt{6}}{4}$
D
none of these

Solution:

$\begin{array}{l} \frac{1}{(\sqrt{2} - \sqrt{3}) + 1} = \frac{1 [(\sqrt{2} - \sqrt{3}) - 1]}{{(\sqrt{2} - \sqrt{3})}^{2} - {(1)}^{2}} = \frac{\sqrt{2} - \sqrt{3} - 1}{2 + 3 - 2 \sqrt{6} - 1} = \frac{\sqrt{2} - \sqrt{3} - 1}{4 - 2 \sqrt{6}} \\ \Rightarrow \frac{(\sqrt{2} - \sqrt{3} - 1) (4 + 2 \sqrt{6})}{{(4)}^{2} - {(2 \sqrt{6})}^{2}} = \frac{(\sqrt{2} - \sqrt{3} - 1) (4 + 2 \sqrt{6})}{16 - 24} \\ \Rightarrow \frac{- 2 (1 + \sqrt{3} - \sqrt{2}) (2 + \sqrt{6})}{- 8} = \frac{(1 + \sqrt{3} - \sqrt{2}) (2 + \sqrt{6})}{4} \end{array}$

$\frac{1}{\sqrt{2} + \sqrt{1} - \sqrt{3}} =$

Solution:

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